43.8. DUALITY AND INTERPOLATION 1463

Lemma 43.8.3 Let f (t) ≥ 0, and let f (t) = α i for t ∈ [2i,2i+1) where α ∈ λθ ,q. Then

there exists a constant, C, such that∣∣∣∣∣∣t−θ f∣∣∣∣∣∣

Lq(0,∞; dtt )≤C ||α||

λθ ,q . (43.8.59)

Also, if whenever α ∈ λθ ,q, and α i ≥ 0 for all i,

∑i

f(2i)2−i

α i ≤C ||α||λ

θ ,q , (43.8.60)

then ∣∣∣∣∣∣{ f(2i)}∞

i=−∞

∣∣∣∣∣∣λ

1−θ ,q′ ≤C. (43.8.61)

Proof: Consider 43.8.59.∫∞

0

(t−θ f (t)

)q dtt= ∑

i

∫ 2i+1

2it−θq

αqi

dtt

≤∑i

∫ 2i+1

2i

(2−iθ

α i

)q dtt= ln2∑

i

(2−iθ

α i

)q= ln2 ||α||q

λθ ,q .

43.8.61 is next. By 43.8.60, whenever α ∈ λθ ,q,∣∣∣∣∣∑i

(f(2i)2−(1−θ)i

)2−θ i

α i

∣∣∣∣∣≤C∣∣∣∣∣∣{2−θ i |α i|

}∣∣∣∣∣∣lq.

It follows from the Riesz representation theorem that{

f(2i)

2−(1−θ)i}

is in lq′ and∣∣∣∣∣∣{ f(2i)2−(1−θ)i

}∣∣∣∣∣∣lq′

=∣∣∣∣{ f

(2i)}∣∣∣∣

λ1−θ ,q′ ≤C.

This proves the lemma.The dual space of (A0,A1)θ ,q,J is discussed next.

Lemma 43.8.4 Let θ ∈ (0,1) and let q≥ 1. Then,

(A0,A1)′θ ,q,J ⊆

(A′1,A

′0)

1−θ ,q′

and the inclusion map is continuous.

Proof: Let a′ ∈ (A0,A1)′θ ,q,J . Now

A0∩A1 ⊆ (A0,A1)θ ,q,J

and ifa ∈ (A0,A1)θ ,q,J ,